Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Calculate:$$\frac{ \left| x \right| }{2}= \frac{1}{x^2+1}$$

How do I write the whole process so it will be correct? I need some suggestions. Thank you!

share|cite|improve this question
There are two cases: (1) $x\geq 0$ and (2) $x<0$. You could consider each separately. – poirot Jan 20 '14 at 16:33
Calculate what?? – Alexey Jan 20 '14 at 22:47
I think you mean "solve for $x$", not "calculate". – Jack M Jan 20 '14 at 23:21
Why is this tagged calculus? – 1110101001 Jan 21 '14 at 4:27
$\large x^{2} = \left\vert\,x\,\right\vert^{2}$ and you get a cubic equation for $\large\left\vert\,x\,\right\vert$. – Felix Marin Jan 23 '14 at 16:48
up vote 15 down vote accepted

Split it into two cases. First case is $$\frac{x}{2}=\frac{1}{x^2+1}.$$ If it has a positive solution (or more than one), it is a valid solution. Second case is $$\frac{-x}{2}=\frac{1}{x^2+1}.$$ If it has a negative solution (or more than one), then it is a valid solution.

The valid solutions from both cases are all your solutions to the original problem.

share|cite|improve this answer
so, you assume $x$ to be real? – lab bhattacharjee Jan 20 '14 at 16:34
I did because of the tags: calculus, and absolute-value. No mention of complex or modulus. – Jeff Snider Jan 20 '14 at 16:58

Let $y = \vert x \rvert \ge 0$. Then you have

$$y(y^2+1)=2 \implies y^3 +y - 2 = 0 \implies (y-1)(y^2+y+2)=0$$

The second factor cannot be zero for non-negative $y$, so you have only one solution $y=1 \implies x = \pm1$.

share|cite|improve this answer
+1 Doing cases $|x| = x$ or $|x| = -x$ is nice, but noting that $|x|^2 = x^2$ saves work at the end checking to make sure $x$ has the right sign. I like this answer best. – 6005 Jan 21 '14 at 3:42

If $x=a+ib$ where $a,b$ are real

We have $$\frac{\sqrt{a^2+b^2}}2\left(a^2-b^2+2abi+1\right)=1$$

Equating the imaginary parts, $\displaystyle ab=0$

If $\displaystyle a=0, \frac{|b|}2(1-b^2)=1$

Use for real $b,|b|=\begin{cases} +b &\mbox{if } b\ge0 \\ -b & \mbox{if } b<0\end{cases}$

If $\displaystyle b=0, \frac{|a|}2(1+a^2)=1$

Set $x=r(\cos\phi+i\sin\phi)$ where $r>0, \phi$ are real

So using de Moivre's formula, $ x^2=r^2(\cos2\phi+i\sin2\phi)$

So we have $\displaystyle \frac r2\left(r^2(\cos2\phi+i\sin2\phi)+1\right)=1$

Equating the imaginary parts, $\displaystyle r^3\sin2\phi=0$

Clearly, $\displaystyle r\ne0\implies \sin2\phi=0\implies2\cos\phi\sin\phi=0$

If $\displaystyle\sin\phi=0, r^3+r-2=0$

Clearly, $r=1$ is a solution, so please solve the Quadratic Equation $\displaystyle\frac{r^3+r-2}{r-1}=0$

If $\displaystyle\cos\phi=0, r^3-r+2=0$ whose solution is not so smooth

share|cite|improve this answer
I got only the following solutions (and none of them are real) $x=\frac{1}{2}(1-i\sqrt{7})$ and $x=\frac{1}{2}(1+i\sqrt{7})$. Is this even correct? – L_McClain Jan 20 '14 at 16:45
@L_McClain, I'm editing my answer – lab bhattacharjee Jan 20 '14 at 16:46
@ lab bhattacharjee, No, it's not that. I calculated on 2 different ways these x-es, and I only get imaginary solutions (?) – L_McClain Jan 20 '14 at 16:48
@L_McClain, how about $x=1$. Let me add the '-' sign – lab bhattacharjee Jan 20 '14 at 16:52
It's IMHO much easier to note that since the LHS is purely real, the RHS must be also, and so $x^2$ must be purely real; this means that either $x=r$ or $x=ri$ for some $r\in\mathbb{R}$, and these two cases can quickly be checked separately. – Steven Stadnicki Jan 20 '14 at 23:42

By "calculuate", I presume you mean "solve for $x$". In that case, simply write two variations of the equation:

$$\frac{ x }{2}=\frac{1}{x^2+1}$$


$$\frac{ -x }{2}=\frac{1}{x^2+1}.$$

The solution set of you're original equation is the union of the solution sets of these two.

share|cite|improve this answer
It is important that you note that in the first equation, x must be non-negative, and x must be negative for the second equation. Otherwise in a more complicated question you may introduce false solutions. – SimonT Jan 21 '14 at 0:03

I'd like to offer a different approach to @labbhattacharjee's attempt to solve the equation over $\mathbb C$.


The left hand side is real, so at minimum the right hand side has to be real. The inverse of a real number is real, so $x^2+1$ is real, and the reals are closed under addition, so $x^2$ is real. Thus $x$ is either purely imaginary or purely real. The latter case has already been solved, and in the prior case, the equation reduces to:


Which can presumably be solved for a real variable in much the same way as the original equation.

share|cite|improve this answer

Here's a plot of $\frac{ \left| x \right| }{2}-\frac{1}{x^2+1}$ in the complex plane (with corners at $-5-5i$ and $5+5i$):

enter image description here

The zeros at $x=\pm 1$ are visible as black sinks. The colorizing is such that modulus is encoded as brightness and phase is encoded as hue. Not meant to be taken as a serious answer, but meant to be a visual supplement.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.